


) ändrar spinnet/polariseringen av en specifik foton . Jag bara filtrerar bort de som INTE uppfyller vertikalpolarisering, eller ? Det var så jag menade iaf. Och polaroider funkar ju så normalt, hmm, fan man börjar bli osäker på allt nu hehe....

Regarding the classical limit, your friend's opinion is certainly simplistic. As regards the hbar -> 0 limit of the Feynman path-integral formulation, here is what I would say if I were to make another edition of my book:Det är lite roligt att trådarna på FB nu börjar involvera utlåtanden från experter.
The Feynman path-integral formulation of QM is just another way of computing the wavefunction. More precisely, it is a way of computing the propagator, which takes the initial wavefunction at time t=0 to the final wavefunction at time t. But all it gives you is a wavefunction, so it does not alleviate the conceptual problem: - How do you get a classical trajectory out of a wavefunction?
As I demonstrated explicitly in my book, and in several papers, the classical (hbar -> 0) limit of a quantum state is, generically, an ensemble of classical trajectories, not a single trajectory. (This is the simplest case - matters might even be more complicated.) This fact is not changed at all if the wavefunction is computed by the Feynman path-integral method.
Let us take a very simple example -- free propagation from a small wavepacket as the initial state. As is obvious, the final wavefunction will retain the spherical symmetry of the initial state. Now the Feynman propagator from a space-time point (X_o, t_o) to the space-time point (X_1, t_1) will be dominated by the single classical path that connects those two space-time points. But the result is still a spherical wavefunction at t = t_1, which does not look at all like a classical trajectory in a definite direction.
The Feynman path-integral method has its uses, but I do not believe that it has any special value in the problem of the classical limit.
As to your question, "Can "hbar -> 0" be regarded as THE LIMIT?", that limit is not defined until you specify what is to be held constant as hbar -> 0. People who make simplistic statements, as did your friend, tacitly assume that you just hold constant all other symbols in the equation except hbar. (Ask him to try it on the Schroedinger equation, where it will give him an absurd result!) Consider, for example, the kinetic energy. If you write it as (hbar*k)^2/(2m) and hold k constant you get an entirely different answer than if you write it as p^2/(2m) with p constant. Believe it or not, I have refereed (and rejected) papers that were submitted for publication, in which the argument turned on such trivialities!
The problem of the classical limit -- better stated as the emergence of classical properties from quantum mechanics -- is much more subtle than many people realize.
Yours truly,
Leslie Ballentine
Regarding the classical limit, your friend's opinion is certainly simplistic. As regards the hbar -> 0 limit of the Feynman path-integral formulation, here is what I would say if I were to make another edition of my book:
The Feynman path-integral formulation of QM is just another way of computing the wavefunction. More precisely, it is a way of computing the propagator, which takes the initial wavefunction at time t=0 to the final wavefunction at time t. But all it gives you is a wavefunction, so it does not alleviate the conceptual problem: - How do you get a classical trajectory out of a wavefunction?
As I demonstrated explicitly in my book, and in several papers, the classical (hbar -> 0) limit of a quantum state is, generically, an ensemble of classical trajectories, not a single trajectory. (This is the simplest case - matters might even be more complicated.) This fact is not changed at all if the wavefunction is computed by the Feynman path-integral method.
Let us take a very simple example -- free propagation from a small wavepacket as the initial state. As is obvious, the final wavefunction will retain the spherical symmetry of the initial state. Now the Feynman propagator from a space-time point (X_o, t_o) to the space-time point (X_1, t_1) will be dominated by the single classical path that connects those two space-time points. But the result is still a spherical wavefunction at t = t_1, which does not look at all like a classical trajectory in a definite direction.
The Feynman path-integral method has its uses, but I do not believe that it has any special value in the problem of the classical limit.
As to your question, "Can "hbar -> 0" be regarded as THE LIMIT?", that limit is not defined until you specify what is to be held constant as hbar -> 0. People who make simplistic statements, as did your friend, tacitly assume that you just hold constant all other symbols in the equation except hbar. (Ask him to try it on the Schroedinger equation, where it will give him an absurd result!) Consider, for example, the kinetic energy. If you write it as (hbar*k)^2/(2m) and hold k constant you get an entirely different answer than if you write it as p^2/(2m) with p constant. Believe it or not, I have refereed (and rejected) papers that were submitted for publication, in which the argument turned on such trivialities!
The problem of the classical limit -- better stated as the emergence of classical properties from quantum mechanics -- is much more subtle than many people realize.
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